2583's Octocanum Modules

[AdamHeard]

Originally Posted by ferrari77

I went through the calculations just now and it seems like it may not be able to lift the robot so we're prepared to switch them out to 1.5'' bore.

I can't figure out how all the forces are acting but looking at how the weight is acting through the mecanum wheel straight up and how the module is pushing sideways it seems like it may roll the wheels outward to switch wheels instead of just pushing straight down to raise the wheel. Would this help at all?

If someone could help me draw a FBD and figure that out, i've highlighted where the piston pushes and the rotation points

Post these same pics with the horizontal and vertical distances between the cylinder point and pivot points noted.

Also x-y distance to wheel shaft from pivot point

[jee7s]

It's a torque thing. Here's a rough verbal sketch of what to do off the top of my head. I"m sure if I sat down and drew it out, I would find an error or nuance. But, here goes...

Draw a vector from the rotation axis of the module to the point where the cylinder applies force. Then, draw a vector from the point where the cylinder applies the force in the direction of the force with the magnitude of the force from the cylinder. Take the cross product of these two vectors. Or, if we make an assumption of planarity and only need the magnitude of the cross product, then Torque = Force * Radius * sin(theta), where theta is the angle between the vectors. This is the "lifting" torque you get from your cylinder.

To determine the torque you need to lift, do the same math, except draw the vector from the rotation axis of the module to the contact point on the ground of the traction wheel. The force then becomes the amount of weight the wheel supports. This is probably pretty close to 1/4 of total robot weight, assuming you have a center of mass close to the center of the robot. The angle between the vectors changes as well, as the weight vector is straight down.

If the torque necessary to lift is less than the torque provided by the cylinder, you aren't getting the mecanum wheels off the ground.

Hope that helps.

[lcoreyl]

make sure to look at the geometry when the colson is just coming off the ground. The pic you posted is looking at after the module has been shifted, and the angles are different.

Also, unless I'm missing something here, the traction position is the more problematic one. the cylinders are also fighting gravity with the colsons down, but now they are "pulling" so they have less force. Also, it appears the geometry here is worse for the cylinders.

Lastly, how did you calculate your top speed reduction?

[ferrari77]

Here are the screenshots of the dimensions in both positions







I used 150 lbs as the robot weight so that gave me 75 lbs on each side and using 1.0625 bore i got (pi*(1.0625/2)^2*60) i got 53 lbs of pushing force for each side, (106 lbs total and 150 lbs of robot weight)

In the first orientation the calculation is 1.3in*75lbs=97.5lbft of torque from the wheel and 3.21in*53lbs=170lbft of torque from the cylinder which provides almost double the force through the cylinder

In the second orientation the calculation is 1.54in*75=115.5lbft of torque from the wheel and 4.1in*53lbs=217.3lbft of torque from the cylinder which provides almost double the force

The mistake i might have made is, because we are using the same cylinder to push two of the modules is the force cut in half so there would actually be 85 and 108lbft of torque, respectively? This would not give us enough torque and would be problematic

The top speed was calculated using a CIM speed of 5300 and a 90% efficiency. It is around 7 fps for the colsons and 20 for the mecanums